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Colligative Properties of Solutions Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

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Colligative Properties n Lowering vapor pressure n Raising boiling point n Lowering freezing point n Generating an osmotic pressure

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Colligative Properties n Lowering vapor pressure n Raising boiling point n Lowering freezing point n Generating an osmotic pressure

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Boiling Point Elevation n a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.

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Boiling Point Elevation n T b = k b m where: T b = elevation of boiling pt m = molality of solute k b = the molal boiling pt elevation constant n k b values are constants; see table 15-4, p. 472 (honors text) n k b for water = 0.52 C/m

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Ex: What is the normal boiling pt of a 2.50 m glucose, C 6 H 12 O 6, solution? n normal implies 1 atm of pressure n T b = k b m n T b = (0.52 C/m)(2.50 m) n T b = 1.3 C n T b = 100.0 C + 1.3 C = 101.3 C

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Ex: How many grams of glucose, C 6 H 12 O 6, would need to be dissolved in 535.5 g of water to produce a solution that boils at 101.5C? n T b = k b m n 1.5 C= (0.52 C/m)(m) n m = 2.885

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Freezing/Melting Point Depression n The freezing point of a solution is always lower than that of the pure solvent.

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Freezing/Melting Point Depression n T f = k f m where: T f = lowering of freezing point m = molality of solute k f = the freezing pt depression constant n k f for water = 1.86 C/m n k f values are constants; see table 15-5, p. 474 (honors text)

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Ex: Calculate the freezing pt of a 2.50 m glucose solution. n T f = k f m n T f = (1.86 C/m)(2.50 m) n T f = 4.65 C n T f = 0.00 C - 4.65 C = -4.65 C

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Ex: When 15.0 g of ethyl alcohol, C 2 H 5 OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20C. The freezing pt of pure formic acid is 8.40C. Determine K f for formic acid. T f = k f m 1.20 C= (k f )( 0.4348 m) k f = 2.8 C/m

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Ex: An antifreeze solution is prepared containing 50.0 cm 3 of ethylene glycol, C 2 H 6 O 2, (d = 1.12 g/cm 3 ), in 50.0 g water. Calculate the freezing point of this 50-50 mixture. Would this antifreeze protect a car in Chicago on a day when the temperature gets as low as 10 F? (-10 F = -23.3 C) T f = k f m T f = (1.86 C/m)(18.06 m) T f = 33.6 C T f = 0 C 33.6 C = -33.6 C YES!

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Electrolytes and Colligative Properties Colligative properties depend on the # of particles present in solution. Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

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Electrolytes and Colligative Properties n For example, the freezing pt of water is lowered by 1.86C with the addition of any molecular solute at a concentration of 1 m. Such as C 6 H 12 O 6, or any other covalent compound n However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72Cdouble that of a molecular solute. NaCl Na + + Cl - (2 particles)

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Electrolytes - Boiling Point Elevation and Freezing Point Depression The relationships are given by the following equations: n T f = k f mn or T b = k b mn T f/b = f.p. depression/elevation of b.p. m = molality of solute k f/b = b.p. elevation/f.p depression constant n = # particles formed from the dissociation of each formula unit of the solute

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Ex: What is the freezing pt of: a) a 1.15 m sodium chloride solution? n NaCl Na + + Cl - n=2 n T f = k f mn n T f = (1.86 C/m)(1.15 m)(2) n T f = 4.28 C n T f = 0.00 C - 4.28 C = -4.28 C

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Ex: What is the freezing pt of: b) a 1.15 m calcium chloride solution? n CaCl 2 Ca 2+ + 2Cl - n=3 n T f = k f mn n T f = (1.86 C/m)(1.15 m)(3) n T f = 6.42 C n T f = 0.00 C 6.42 C = -6.42 C

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Ex: What is the freezing pt of: c) a 1.15 m calcium phosphate solution? n Ca 3 (PO 4 ) 2 3Ca 2+ + 2PO 4 3- n n=5 n T f = k f mn n T f = (1.86 C/m)(1.15 m)(5) n T f = 10.7 C n T f = 0.0 C 10.7 C = -10.7 C

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Determining Molecular Weights by Freezing Point Depression

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n T f = 0.56C n T f = k f m n 0.56C = (5.12C/m)(m) n m = 0.1094 Ex: A 1.20 g sample of an unknown molecular compound is dissolved in 50.0 g of benzene. The solution freezes at 4.92C. Determine the molecular weight of the compound. The freezing pt of pure benzene is 5.48C and the K f for benzene is 5.12C/m.

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Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0 g of water. The resulting solution froze at 5.58C. What is the molecular weight of the compound? n T f = 5.58C n T f = k f m n 5.58C = (1.86C/m)(m) n m = 3.00 m